Problem 71 - Algebra for College Students

Question

Simplify each of the following. Express final results using positive exponents only. For example,\(\left(2 x^{\frac{1}{2}}\right)\left(3 x^{\frac{1}{3}}\right)=6 x^{\frac{5}{6}}\). \(\frac{48 b^{\frac{1}{3}}}{12 b^{\frac{3}{4}}}\)

Step-by-Step Solution

Answer

The simplified expression is \( \frac{4}{b^{\frac{5}{12}}} \).

Step 1: Simplify the Numerical Coefficient

First, look at the numerical coefficients in the fraction \( \frac{48}{12} \). Divide 48 by 12 to simplify it to \( 4 \). This means that the numerical part of our simplified expression is \( 4 \).

Step 2: Apply the Law of Exponents

Using the law of exponents \( \frac{a^m}{a^n} = a^{m-n} \), simplify the expression \( \frac{b^{\frac{1}{3}}}{b^{\frac{3}{4}}} \). Calculate the new exponent by subtracting the exponents: \( \frac{1}{3} - \frac{3}{4} \).

Step 3: Find a Common Denominator

To subtract \( \frac{1}{3} \) and \( \frac{3}{4} \), find a common denominator for the fractions. The lowest common denominator for 3 and 4 is 12. Convert the fractions: \( \frac{1}{3} = \frac{4}{12} \) and \( \frac{3}{4} = \frac{9}{12} \).

Step 4: Subtract the Exponents

Subtract the two fractions: \( \frac{4}{12} - \frac{9}{12} = -\frac{5}{12} \). This means the exponent of \( b \) in the simplified expression is \( b^{-\frac{5}{12}} \).

Step 5: Express Exponent as Positive

Since we need to express with positive exponents, use \( a^{-n} = \frac{1}{a^n} \) to rewrite \( b^{-\frac{5}{12}} \) as \( \frac{1}{b^{\frac{5}{12}}} \).

Step 6: Combine the Simplified Parts

Now combine the simplified numeric coefficient and variable part to get the final expression: \[ \frac{4}{b^{\frac{5}{12}}} \].

Key Concepts

Exponents

Exponents are a way of denoting repeated multiplication of a number by itself. For example, the expression \( a^3 \) means that the number \( a \) is multiplied by itself three times. Exponents provide a shorthand method of representing long multiplication processes.
A few things to remember about exponents include:

• An exponent of zero means any non-zero base raised to this power equals one (e.g., \( a^0 = 1 \)).
• Negative exponents represent the reciprocal of the base raised to the corresponding positive exponent (e.g., \( a^{-n} = \frac{1}{a^n} \)).

In the expression \( rac{48 b^{\frac{1}{3}}}{12 b^{\frac{3}{4}}} \), exponents are used to indicate parts of a quantity that repeat less than once, as seen in fractional exponents.

Fraction Simplification

Fraction simplification is the process of reducing fractions to their simplest form. This means dividing the numerator and the denominator by their greatest common divisor until no further simplification is possible.
Consider the numbers \( 48 \) and \( 12 \) in our example. Both numbers can be divided by \( 12 \), leading to a simplification: \( \frac{48}{12} = 4 \).
This step is crucial when working with fractions that include variables raised to fractional exponents. It helps ease further operations by breaking down the expression into its basic numerical components.
Once simplification of the numbers is complete, we simplify the exponents separately, which we will cover under the law of exponents.

Law of Exponents

The laws of exponents are a set of rules that govern operations on expressions with exponents. They assist in simplifying expressions effectively. The key laws include:

• Product of powers: When multiplying two exponents with the same base, add the exponents (\( a^m \times a^n = a^{m+n} \)).
• Quotient of powers: When dividing two exponents with the same base, subtract the exponents (\( \frac{a^m}{a^n} = a^{m-n} \)).
• Power of a power: When raising an exponent to another power, multiply the exponents (\((a^m)^n = a^{m\times n}\)).

In our original example, we used the quotient of powers to simplify the exponents: \( \frac{b^{\frac{1}{3}}}{b^{\frac{3}{4}}} = b^{\frac{1}{3} - \frac{3}{4}} = b^{-\frac{5}{12}} \). Finally, it is important to express this using a positive exponent, converting the result to \( \frac{1}{b^{\frac{5}{12}}} \). This makes the expression easier to interpret and work with in mathematical operations.